## Wednesday, October 18, 2017

### Lesson 4-4: The First Theorem in Euclid's Elements (Day 44)

This is what Theoni Pappas writes on page 291 of her Magic of Mathematics:

"I make no question but you will readily allow the square of 16 to be the most magically magical of any magic square ever made by a magician."
-- Benjamin Franklin

This is the first page of a new section, "Magic Squares & Other Recreations." And this subsection, which spans several pages, is called "Playing with Magic Squares."

You may already be aware of the concept of a magic square. This is usually defined as a square matrix whose entries are distinct, yet whose rows, columns, and diagonals have the same sum. The smallest possible magic square has dimension (or order) 3, and Pappas provides a trio of examples:

8 1 6
3 5 7
4 9 2

120 15   90
45   75   105
60   135 30

124 19   94
49   79   109
64   139 34

Why exactly do we call these "magic squares" anyway? Pappas explains:

"Imagine the formation of numbers into a square with special manipulative powers."

In other words, people were superstitious and believed that the square was actually magic. Pappas explains how the ancient Chinese and medieval Egyptians sought divine powers from the square. And Pappas even writes:

"In the 8th century A.D., certain alchemists believed that the these squares held the key to converting metals to gold."

And as we see by the opening quote, Founding Father Ben Franklin also believed in its magic. We will see more magic squares, including Franklin's square of order 16 (not 4, but 16, as in 16 * 16) in the next few posts.

Meanwhile, let's give the answer to yesterday's question. The longest side (length 27) is parallel to a medium side (length 18). Draw as a dotted line the perpendicular bisector of this medium side, and note where this dotted line meets the longest side. From this new point, we draw two solid lines to represent the cut lines. The two cut lines are perpendicular lines both starting at this point -- one of them ends where two medium sides meet and the other ends where two short sides (length 9) meet.

After cutting, we obtain two right triangles of legs 9 and 18, and a large concave pentagon. The two triangles readily fit inside the concave pentagon to form the square.

Notice that it doesn't matter that the original side lengths are multiples of nine. We could just as well have made the side lengths 1, 2, and 3. In any case, the cut lines are diagonals of rectangles of dimension 2 * 1, which makes sense because we already determined in the previous post that the length of the final square is a multiple of sqrt(5), the diagonal of a 2 * 1 rectangle.

Chapter 4 of Paul Hoffman's The Man Who Loved Only Numbers is "Marginal Revenge." As usual, the chapter begins with a quote:

Thomasina: If you do not teach me the true meaning of things, who will?
Septimus:....

Oh, so now we begin with lines from a play, Tom Stoppard's Arcadia, Act I, Scene I. Well, let's skip over the part where Septimus talks about sex and skip to the juicy part:

Septimus: Fermat's Last Theorem, by contrast, asserts that when x, y, and z are whole numbers each raised to power of n, the sum of the first two can never equal the third when n is greater than 2. (Pause.)
Thomasina: Eurghhh!

Obviously, that's the joke -- Septimus and Thomasina are more disgusted by FLT (which, of course, we've already discussed in Ogilvy) than by sex. Paul Erdos, of course, feels the exact opposite.

In my last two posts, I rushed through two chapters each day, and I feel that therefore, I didn't do those chapters justice -- especially Chapters e and pi. Let's see if I can cover today's single Chapter 4 in fuller detail.

Hoffman begins with a story about baseball -- in 1974, Hank Aaron hit his 715th home run to break Babe Ruth's career record of 714. Yes, I know that the baseball playoffs are heating up as we're close to determining the two World Series participants, but that's not why I mention baseball here. Instead, Hoffman writes about Carl Pomerance, a mathematician who works at the University of Georgia, a few miles from where Aaron hit the historic homer. (He is still alive at the age of 73.) Pomerance notes that the numbers 714 and 715 aren't just home run totals but in fact have a special property:

714 * 715 = 2 * 3 * 5 * 7 * 11 * 13 * 17

So their product is the same as that of the first seven primes. Furthermore, let's actually factor 714 and 715 separately, and then add up the prime factors:

714 = 2 * 3 * 7 * 17
715 = 5 * 11 * 13
2 + 3 + 7 + 17 = 5 + 11 + 13

The sums of their prime factors are equal. And so Pomerance defines a Ruth-Aaron pair as a pair of consecutive integers whose sums of their prime factors are equal. He uses a computer to find 26 such Ruth-Aaron pairs, from 5 and 6 to 18,490 and 18,491 He makes a conjecture that there exist infinitely many Ruth-Aaron pairs, and this conjecture is proved a week later -- by Paul Erdos, of course. And according to Hoffman, Erdos, Pomerance, and Hank Aaron meet in 1995. Erdos and Aaron sign a baseball for Pomerance, who then proclaims, "And thus Hank Aaron has Erdos number one."

Pomerance, of course, himself has Erdos number 1, as the two mathematicians collaborate on many papers relating to Fermat's Little Theorem. I've written about this theorem in the past, as it's used to determine whether a number is prime. From Ogilvy, we already know that if n is prime, then by Fermat's Little Theorem, a^n - a is a multiple of n for each a. But he also tells us that the converse is false, as 2^341 - 2 is a multiple of 341, yet 341 = 31 * 11 isn't prime. So 341 is a base-2 pseudoprime.

Now Hoffman goes one step further and tells us that there exist some composite numbers n such that a^n - a is a multiple of n not just when a = 2, but for every value of a. Such numbers are called Carmichael numbers, named for their discoverer, Robert Carmichael. He found out that 561, which equals 3 * 11 * 17, is indeed pseudoprime in every base. Ultimately, Erdos and Pomerance prove that there are infinitely many Carmichael numbers.

The title of this chapter mentions a "margin" -- referring to Fermat's margin. But it wasn't his Little Theorem that appeared in this margin (and in the opening quote) -- it was his Last Theorem. And so Hoffman devotes the remainder of this chapter to FLT and the mathematicians who proved various cases of FLT, beginning with Fermat himself and ending up with Andrew Wiles.

Hoffman describes Pierre de Fermat as a 17th century French mathematician. Fermat was a judge who served King Louis XIV. He was encouraged to spend his evenings alone in order to avoid potential conflicts of interest in court. And so he spent his spare time reading about -- what else would it be -- mathematics. In particular, he read the works of the ancient Greek mathematician Diophantus -- recall that Ogilvy devotes an entire chapter to Diophantine equations.

Hoffman tells us a commonly told story about Diophantus, where we must actually solve a Diophantine equation to find the details of his life. Here's a link to the story:

https://www.mathsisfun.com/puzzles/diophantus.html

"This tomb holds Diophantus. Ah, what a marvel! And the tomb tells scientifically the measure of his life. God vouchsafed that he should be a boy for the sixth part of his life; when a twelfth was added, his cheeks acquired a beard; He kindled for him the light of marriage after a seventh, and in the fifth year after his marriage He granted him a son. Alas! late-begotten and miserable child, when he had reached the measure of half his father's life, the chill grave took him. After consoling his grief by this science of numbers for four years, he reached the end of his life."

The link above explains why Diophantus lived to be 84-- a ripe old age considering that he lived in the third century AD. The equation to solve is:

D = D/6 + D/12 + D/7  + 5 + D/2 + 4

According to Hoffman, this Diophantine equation  is of the first-degree (or linear). An example of a second-degree (or quadratic) Diophantine equation is the Pythagorean Theorem, which has infinitely many solutions. As Fermat read Diophantus, he is motivated to generalize the Pythagorean Theorem to higher exponents. Back in Ogilvy, we learn of Fermat's proof of the case n = 4. This is most likely the only case the French mathematician ever actually solved. The evidence that he had a general proof is, to say the least, marginal. (Okay, that's a bad pun!) Later on in the chapter, Hoffman tells us that Fermat was probably mistaken that he had a proof, just as he was mistaken that every number of the form 2^2^n + 1 is prime (the primes named for him), when in the end, it appears that only the first five of them (from n = 0 to n = 4) are.

The next case to be proved was n = 3, by Euler. This is the same Euler for whom the constant e is named, and I also mentioned Euler on the blog in covering Lesson 1-4 of the U of Chicago text. So this left mathematicians with needing to prove the cases for:

x^5 + y^5 = z^5
x^6 + y^6 = z^6
x^7 + y^7 = z^7
x^8 + y^8 = z^8

and so on.

Hoffman tells us that the next person to prove a case of FLT was the 18th-century French mathematician Sophie Germain. After glossing quickly over Euler, Hoffman devotes several pages to
Germain's story.

As it happened, Germain, like her countryman Fermat, discovered her passion for math when she was reading while secluded in her home -- in this case to escape the violence of the French Revolution. In fact, she was only 13 when she read about Archimedes and his final words, spoken to the Roman soldiers who were about to kill him -- "Don't step on my circles!" (I've mentioned this story of Archimedes on the blog in April 2016.) This fascinated the young teenage girl, who thought that math was so important to Archimedes even during the dark times of war, she should study it as well during the darkness of the Reign of Terror.

When Germain was 18, the famous Ecole Polytechnique opened -- it was soon to become the school for famous mathematicians. As it turned out, Germain was barred from attending due to her gender, yet she managed to obtain lecture notes. She learned about FLT by reading the lecture notes. In the end, she doesn't prove FLT for just one exponent, but for infinitely many exponents -- specifically the primes p such that 2p + 1 is also prime. These primes are now called Sophie Germain primes in her honor, and even though FLT has now been proved for all cases, Sophie Germain primes are an interesting topic. Like twin primes, it's still unknown whether there are infinitely many Sophie Germain primes. The largest known such prime at the time of Hoffman is 92,305 * 2^16,998 + 1, and in 2017, the largest known Sophie Germain prime is 2,618,163,402,417 * 2^1,290,000 - 1.

Germain sends her partial FLT proof to the famous mathematician Gauss, under a male pseudonym in order to be taken seriously. He finally learns of her identity when she sends a letter on his behalf to an army commander in order to protect the famous German mathematician from Napoleon.

The next mathematician to prove cases of FLT was a German, Ernst Eduard Kummer. He proves FLT for several primes called "regular primes." Hoffman tells us that unlike Sophie German primes, regular primes are too difficult to define in his book.

The author tells us that Kummer, with all of his vast math knowledge, may be a "dren." When teaching one day, he has trouble multiplying 7 * 9. To get this back to Erdos, Hoffman tells us a joke Erdos makes about Kummer -- he multiplies 7 * 9 by eliminating 61 and 67 as they're prime, then 65 as it's a multiple of 5, and then 69 as it's too big. The only odd number left in the 60's is 63.

When Hoffman finally reaches Wiles himself, he tells us that the brilliant British mathematician first reads about FLT when he is only ten years old, in a book called The Last Problem by E.T. Bell. As soon as the young lad learn of the problem, he is surprised that a question so easily stated is so difficult to solve. He makes it his lifelong mission to prove the theorem -- and he finally does.

Hoffman writes about how Wiles gives a surprise lecture on FLT in 1993. But this proof has an error, which he doesn't fix until the following year. Pappas writes her book during the time when Wiles is working on correcting the error, which is why she's able to mention Wiles in her book, but with a little uncertainty that the proof is correct.

OK, I've written so much about Erdos and Wiles, Germain and Gauss, Euler and Fermat. But let's discuss the one who comes before all of them, including Archimedes -- Euclid.

Lesson 4-4 of the U of Chicago text is called "The First Theorem in Euclid's Elements." This lesson from the Second Edition, like yesterday's 4-3, has no exact counterpart in the Third Edition. The closest corresponding chapter would be part of Lesson 5-4.

This is what I wrote two years ago about today's lesson:

Lesson 4-4 of the U of Chicago text introduces formal, two-column proofs. As a first example of a proof, the text gives Proposition 1 from Euclid. Since Euclid was among the first to write formal proofs and this was his first theorem, the students' first proof will be one of the oldest proofs written in the whole world.

I already mentioned David Joyce's website and his Euclid pages. Here's a link to Proposition 1:

http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI1.html

And here's what Joyce writes about why Euclid chose this one to be first:

"This proposition is a very pleasant choice for the first proposition in the Elements. The construction of the triangle is clear, and the proof that it is an equilateral triangle is evident. Of course, there are two choices for the point C, but either one will do.

"Euclid could have chosen proposition I.4 [SAS Congruence -- dw] to come first, since it doesn’t logically depend on the previous three, but there are some good reasons for putting I.1 first. For one thing, the Elements ends with constructions of the five regular solids in Book XIII, so it is a nice aesthetic touch to begin with the construction of a regular triangle. More important, though, is I.1 is needed in I.2, and that in I.3. Propositions I.2 and I.3 give constructions for moving lines, and I.4, although not logically dependent on I.2 or I.3, does use the concept of superposition which involves, in some sense, moving points and lines."

The U of Chicago text gives a two-column proof of Euclid's Proposition 1. There are a few differences between a two-column proof in this text and those found in most other texts. First, the U of Chicago labels the two columns "Conclusions" and "Justifications" -- whereas most other texts label them "Statements" and "Reasons." Also, most books start their two-column proofs with the "Given" information, but the U of Chicago skips the "Given" lines. On this blog, I plan on doing proofs the way most traditional texts do them, with "Statements," "Reasons," and "Given."

Notice that this theorem is truly a construction -- if available, teachers can have the students use a compass to draw the circles and a straightedge to draw the segments. This sounds like something mentioned in the Common Core Standards:

CCSS.MATH.CONTENT.HSG.CO.D.13
Construct an equilateral triangle, a square, and a regular hexagon inscribed in a circle.

The question is, must the equilateral triangle and the square be inscribed in the circle, or must only the regular hexagon be so inscribed? After all, Euclid's equilateral triangle is not inscribed in the circle -- for a triangle to be inscribed, all three vertices must lie on the same circle, but the vertices of Euclid's triangle lie on different circles. Unfortunately, the standards are vague here.

Now Example 1 in the text gives another example of a two-column proof. Many books refer to this as the Alternate Exterior (not Interior) Angle Theorem. But we're saving proofs based on parallel lines until a little later. The midpoint proof in Example 2, of course, can be given.

I decided that this is a good time to include two of the proofs that I gave last year -- the Uniqueness of Perpendiculars Theorem and the Line Perpendicular to Mirror Theorem. The first theorem is actually Euclid's Proposition 12:

http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI12.html

I chose these two theorems because they fit here the best. After all, Joyce points out that the same double-equilateral-triangle construction is used here as in Proposition 1. Also, Example 2 in Lesson 4-1 hints at this same proof. The difference between our proof and Euclid's is that we explicitly mention reflections in our proof, since Common Core demands that we use reflections in proof. The Line Perpendicular to Mirror Theorem follows directly from the Uniqueness of Perpendiculars Theorem, and tells us that a line perpendicular to the reflecting line (the mirror) is invariant -- that is, its image is identical to the preimage. Finally, these two theorems on perpendiculars provide a great segue into Lesson 4-5, the Perpendicular Bisector Theorem.

Now let's do the reflections. The key is that the graph of x = h is a line perpendicular to the x-axis for any value of h, and likewise y = k is perpendicular to the y-axis for any value of k. We then reflect the point (hk) by finding the images of the lines x = h and y =k. For example, if the mirror is the y-axis, then any line of the form y = k must be mapped to itself -- this is because y = k is perpendicular to the y-axis, and so by the Line Perpendicular to Mirror Theorem, y = k is an invariant line. That the image of x = h must be x = -h follows from the Reflection Postulate and Ruler Postulate applied to the x-axis.

I'm not sure whether I want to post these proofs for the students yet -- not because the proofs are hard, but because they have enough to worry about as they first learn about proofs. The cases where the mirror is either of the axes can be proved using today's theorems. The mirrors y = x and y = -x can wait until the first part of Chapter 5, where we prove properties of isosceles triangles and kites, but still before we give a parallel postulate. The mirrors x = h and y = k require a parallel postulate, but we only need to consider the properties of rectangles, not similar triangles. No other mirrors appear on the Common Core test -- thank goodness for that!

Regarding the worksheets, as I mentioned earlier, I like the idea of having the students experiment with each theorem by folding the paper (for reflections) before actually proving the theorem. In some ways, this is the Michael Serra approach, except that the proof is given immediately after -- not near the end of the book. Because of this, I'm including several worksheets today. (Don't say I didn't warn you!) I have the statement of each theorem on one side of the page -- large enough to encourage folding -- and a two-column proof on the other, with a few Reasons left blank for the students to write in.

And as for the Exercises, notice that Question 12 contains a flow proof. Many other geometry texts emphasize flow proof as a third type of proof, after paragraph and two-column proofs. But in the U of Chicago texts, a flow proof appears in this question and then never again in the text. Since it's a rewriting of the Alternate Exterior Angle Theorem from Example 1, I'm throwing it out. Let's not confuse the students with flow proofs just yet -- and I'll probably leave flow proof out altogether.

Question 24 is the Exploration/Bonus Question. It directs students to discover that the altitudes of a triangle are congruent -- that is, that they meet at a point (called the orthocenter). Notice that many texts cover the four concurrency proofs for triangles (centroid, circumcenter, incenter, and orthocenter), but these aren't covered fully in the U of Chicago text. The orthocenter never appears in the text again after this Exploration Question. Actually, one concurrency question appears in the Common Core Standards:

CCSS.MATH.CONTENT.HSG.CO.C.10
Prove theorems about triangles. Theorems include ... the medians of a triangle meet at a point.

The proof that the medians meet at the centroid requires similar triangles, and so the proof must wait until the second semester. But the proof that the perpendicular bisectors meet at the circumcenter is the easiest of the concurrency proofs, and it appears in the very next section, 4-5. Incidentally, the proof that the altitudes meet at a point entails constructing a larger triangle and showing that the altitudes of the smaller triangle extend to the perpendicular bisectors of the larger triangle (so the orthocenter of the smaller triangle is the circumcenter of the larger triangle)! But this proof requires parallel lines and Playfair, so it must wait. Still, the students can still explore this in the Exercises.